What is the general solution for 2sin(3θ)=-1 ?

The general solution for 2sin(3θ)=-1 is θ=(7pi}{18}+(2pin)/3 ,θ=(11pi)/(18)+\frac{2pin)/3

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2sin(3θ)=-1

2 sin (3 θ) = - 1

θ = \frac{7 π}{18} + \frac{2 πn}{3}, θ = \frac{11 π}{18} + \frac{2 πn}{3}

Solution steps

2 sin (3 θ) = - 1

Divide both sides by

2

sin (3 θ) = - \frac{1}{2}

General solutions for

sin (3 θ) = - \frac{1}{2}

3 θ = \frac{7 π}{6} + 2 πn, 3 θ = \frac{11 π}{6} + 2 πn

Solve

3 θ = \frac{7 π}{6} + 2 πn : θ = \frac{7 π}{18} + \frac{2 πn}{3}

Solve

3 θ = \frac{11 π}{6} + 2 πn : θ = \frac{11 π}{18} + \frac{2 πn}{3}

θ = \frac{7 π}{18} + \frac{2 πn}{3}, θ = \frac{11 π}{18} + \frac{2 πn}{3}

2 + 2 cos (θ) = 2 - 2 cos (θ)

1 + 4 sin (x) = 0

cos^{2} (x) = \frac{3}{2} sin (x)

sin (x) = cos (\frac{x}{2})

cos (\frac{π ( x - 7 )}{3}) = \frac{1}{2}

What is the general solution for 2sin(3θ)=-1 ?
The general solution for 2sin(3θ)=-1 is θ=(7pi}{18}+(2pin)/3 ,θ=(11pi)/(18)+\frac{2pin)/3